Let $f\left(x\right)$ be a polynomial of degree $3$ such that $f\left(3\right)=1,f\text{'}\left(3\right)=-1,f\text{'}\text{'}\left(3\right)=0,f\text{'}\text{'}\text{'}\left(3\right)=12.$ Then the value of $f\text{'}\left(1\right)$ is

$f\left(x\right)=a(x-3{)}^3+b(x-3{)}^2+c(x-3)+d$

$\begin{array}{r}f\left(3\right)=d=1\end{array}$

$f^{\prime \prime }\left(x\right)=6a(x-3)+2b$

$\begin{array}{l}{f}^1\left(3\right)=3a(x-3{)}^2+2b(x-3)+c\\ f^1\left(3\right)=-1=c\end{array}$

$\begin{array}{l}{f}^{\text{'}\text{'}}\left(3\right)=0\Rightarrow 2b=0\Rightarrow b=0\\ f^{\text{'}\text{'}\text{'}}\left(x\right)=6a\end{array}$

$f\text{'}\text{'}\text{'}\left(3\right)=6a=12\Rightarrow a=2$

$\begin{array}{r}f\left(x\right)=a(x-3{)}^3+b(x-3{)}^2+c(x-3)+d\\ =2(x-3{)}^3+(-1\left)\right(x-3)+1\end{array}$

$f^{\mathrm{\prime }}\left(x\right)=6(x-3{)}^2-1$

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(1\right)=6(1-3{)}^2-1\\ =6\left(4\right)-1=23\end{array}$